A veterinary Nutritionist developed a diet for overweight dogs. The total volume

Question

A veterinary Nutritionist developed a diet for overweight dogs. The total volume of food consumed remains the same, but one-half of the dog food is replaced with a low-calorie "filler" such as canned green beans. Ten overweight dogs were randomly selected from her practice and were put on this program. Their initial weights were recorded, and then they were weighed again after 4 weeks. At the 0.05 level of significance, can it be concluded that the dogs lost weight? Assume the variables are both normally Distributed.

State the hypotheses and identify the clain with the correct Hypothesis

H0

H1

This hypothesis test is a _____ test

Find the critical value(s). Round the answer to three decimal places.

Critical value is _____

Compute the test value. Round intermediate calculations and final answers to three decimal places.

t = _____

Make the decision

______ the null hypothesis

Summarize the results

There __ enough evidence to support the claim that the dogs lost weight.

Before 53 48 65 36 40 52 45 50 39 42 After 45 40 58 34 42 47 41 47 38 39

Explanation / Answer

Given that,
population mean(u)=65
sample mean, x =36
standard deviation, s =40
number (n)=52
null, H0: Ud < 0
alternate, H1: Ud > 0
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.833
since our test is right-tailed
reject Ho, if to > 1.833
we use Test Statistic
to= d/ (S/n)
Where
Value of S^2 = [ di^2 – ( di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = 3.9
We have d = 3.9
Pooled variance = Calculate value of Sd= S^2 = Sqrt [ 245-(39^2/10 ] / 9 = 3.21
to = d/ (S/n) = 3.84
Critical Value
The Value of |t | with n-1 = 9 d.f is 1.833
We got |t o| = 3.84 & |t | =1.833
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
p-value :right tail - Ha : ( p > 3.8386 ) = 0.00199
hence value of p0.05 > 0.00199,here we reject Ho
ANSWERS
---------------
null, H0: Ud < 0
alternate, H1: Ud > 0
test statistic: 3.84
critical value: reject Ho, if to > 1.833
decision: Reject Ho
p-value: 0.00199

X Y X-Y (X-Y)^2 53 45 8 64 48 40 8 64 65 58 7 49 36 34 2 4 40 42 -2 4 52 47 5 25 45 41 4 16 50 47 3 9 39 38 1 1 42 39 3 9 39 245

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